A view through the walls of our classroom. This is an interactive learning ecology for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

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Friday, October 17, 2008

Friday's class of a couple Theroms.

Today in class we learnt about a couple new theorems. The first one is the "Intermediate Value Theorem". This theorem pretty much states that according to slide 1, found on the graph, that if y= f(x) is continuous on an interval [a, b] and y = k in between points y=f(b) and y=f(a) then there is an x-value of c where f(c) = k. If a discontinuity is found, remove it. I believe that what Mrs. E means is if there is a discontinuity between the chosen interval of [a,b] then chose a new interval by moving it to a place on the graph where there is no discontinuity. Confusing? Yeah, just look at the graph and read the notes a couple times. That's what I did.

We also looked at how to find zeros on a continuous function without our calculators. We were given an example on slide 2. f(x) = x3-3x+1. To find the zero of this function, we chose points [0,1] for this example. We then found out where the x-values were either positive or negative. On the example the point 0 was input into the function and equaled to 1. Which was positive ( f(0) = 0-3(0) +1 = 1). At the point of 1, the function was a equal to a negative number ( f(1) = 1 -3(1) + 1 = -1). Now we know that the zero of the function will be in between those two points. Now we want to hone in on a smaller region, so we should cut that interval of 0 to 1, in half. So we find the f(1/2) and see if it's positive or negative, so we know where to look next. The function of f(1/2) turned out to be negative, as found on the slide. We then know to look between points [0 and 1/2] because the zero will be in between a negative and positive value. It's best to divide the section by half to cover greater amount of space between the points. The next point we used was 1/4. We input this into the function and found it to be positive. Now we can look between points [1/4 and 1/2]. We then used the point of 3/8 in our function, which is in between our given points. It turned out this was negative, so now we know our zero is between 1/4 and 3/8 which is small enough of an interval where our zero is found, so we stopped there.

The next theorem we used was entitled the "Extreme Value Theorem". This states that on a continuous function, it will always have a maximum and minimum and will be an extremum value from where all points on the function will either larger or smaller depending if its the maximum or minimum extremum. Lets say that on a continuous function on interval [a,b] there exists numbers c and d where all x-values in [a,b] f(c) ≤ f(x) ≤ f(d). d will be the maximum and all x-values will be less that or equal to d. c is the minimum and all x-values will be greater that c or equal.

That's all we did in today's Calculus class. If anything is wrong, talk to me and I'll be sure to change it. The next scribe will be Lawrence!